设数列{bn}的前项和为sn,且bn=2-2sn：数列{an}为等差数列,且a5=14.a7=20.
2个回答

n+1=2-Sn+1

bn=2-Sn

相减有bn+1-bn=-bn+1,bn+1=bn/2

b1=2-2b1得到b1=2/3,bn=1/3*(1/2)^n;

a7-a5=2d=6,d=3,a1=a5-4d=2; an=3*n-1;

Tn =a1*b1+a2*b2+a3*b3+...+an*bn

1/2*Tn= a1*b2+a2*b3+...+an-1*bn+an*bn+1

相减有1/2*Tn=a1*b1+d*(b2+b3+...+bn)-an*bn+1

=4/3+2*1/3*[1-(1/2)^(n-1)]*2-1/3*(3*n-1)*(1/2)^(n+1)

相关问题